# What is the slope of the tangent line of  (x/y-2)(xy-3)-e^y= C , where C is an arbitrary constant, at (-2,1)?

The slope of the tangent line $m = \frac{- 9}{e + 2} = - 1.90747$

#### Explanation:

We start from the given equation

$\left(\frac{x}{y} - 2\right) \left(x y - 3\right) - {e}^{y} = C$

Expand the left side of the equation

$\left(\frac{x}{y}\right) \left(x y\right) - \frac{3 x}{y} - 2 x y + 6 - {e}^{y} = C$

${x}^{2} - \frac{3 x}{y} - 2 x y + 6 - {e}^{y} = C$

Differentiate both sides of the equation with respect to $x$

$\frac{d}{\mathrm{dx}} \left({x}^{2} - \frac{3 x}{y} - 2 x y + 6 - {e}^{y}\right) = \frac{d}{\mathrm{dx}} \left(C\right)$

$\frac{d}{\mathrm{dx}} \left({x}^{2}\right) - 3 \cdot \frac{d}{\mathrm{dx}} \left(\frac{x}{y}\right) - 2 \cdot \frac{d}{\mathrm{dx}} \left(x y\right) + \frac{d}{\mathrm{dx}} \left(6\right) - \frac{d}{\mathrm{dx}} \left({e}^{y}\right) = 0$

$2 x - 3 \left(\frac{y \cdot 1 - x \cdot y '}{y} ^ 2\right) - 2 \cdot \left(x y ' + y \cdot 1\right) + 0 - {e}^{y} \cdot y ' = 0$

Substitute right away the value of $x = - 2$ and $y = 1$ from the point $\left(- 2 , 1\right)$

$2 x - 3 \left(\frac{y \cdot 1 - x \cdot y '}{y} ^ 2\right) - 2 \cdot \left(x y ' + y \cdot 1\right) + 0 - {e}^{y} \cdot y ' = 0$

$2 \left(- 2\right) - 3 \left(\frac{1 \cdot 1 - \left(- 2\right) \cdot y '}{1} ^ 2\right) - 2 \left[\left(- 2\right) y ' + 1 \cdot 1\right] + 0 - {e}^{1} \cdot y ' = 0$

Simplify

$2 \left(- 2\right) - 3 \left(\frac{1 \cdot 1 - \left(- 2\right) \cdot y '}{1} ^ 2\right) - 2 \left[\left(- 2\right) y ' + 1 \left(1\right)\right] + 0 - \left({e}^{1}\right) y ' = 0$

$- 4 - 3 \left(\frac{1 + 2 y '}{1}\right) + 4 y ' - 2 + 0 - \left({e}^{1}\right) y ' = 0$

$- 4 - 3 - 6 y ' + 4 y ' - 2 - e y ' = 0$

Solve for $y '$

$- 6 y ' + 4 y ' - e y ' = 4 + 3 + 2$

$\left(- 6 + 4 - e\right) y ' = 9$

$y ' = \frac{9}{- 6 + 4 - e}$

$y ' = \frac{9}{- 2 - e}$

$y ' = \frac{- 9}{2 + e}$

$y ' = - 1.90747$

God bless....I hope the explanation is useful.